# Moment Inertia of a cube

1. Jan 12, 2010

### Cosmossos

To which of the two cubes has a larger moment of inertia?

I think it's the right one, is it correct?
How can I explain that without using the parallel axis theorem?

2. Jan 12, 2010

### Staff: Mentor

Why do you say the right one? Are you familiar with the relevant equation for calculating the moment of inertia?

3. Jan 12, 2010

### Cosmossos

what relevant equation?

I think it's the right one because We know that the minimal moment of inertia is throw the principal axes that goes throw the center of mass. in the right one , the rotation isn't throw the principal axes . there is also the following theorem :

The moment of inertia about an arbitrary axis is equal to the
moment of inertia about a parallel axis passing through the
center of mass plus the moment of inertia of the body about
the arbitrary axis, taken as if all of the mass M of the body
were at the center of mass.

Am I wrong?

Last edited: Jan 12, 2010
4. Jan 12, 2010

### Staff: Mentor

There may be a shortcut way to tell which has a higher moment of inertia, but for me, I'd need to calculate it. I'd use the standard definition of the Mmoment of inertia, and evaluate thge integral for the diagonal case. I don't think you can use the parallel axis theorm, since the two axes are not parallel.

I'd do the 2-D case first, to see if it offered some intuition. That is, the moment of inertia for a flat rectangular sheet, with the axes going straight versus diagonal.

5. Jan 12, 2010

### vela

Staff Emeritus
I think they'll turn out to be equal. Try computing the moment of inertia tensor.